What is the integral from 1 to e of x^3*ln(x) ?

The integral from 1 to e of x^3*ln(x) is (3e^4+1)/(16)

integral from 1 to e of x^3*ln(x)

\int_{1}^{e} x^{3} \cdot ln (x) d x

\frac{3 e ^{4} + 1}{16}

Decimal

10.29965 \dots

Solution steps

\int_{1}^{e} x^{3} ln (x) d x

Apply Integration By Parts

= [\frac{1}{4} (x^{4} ln (x) - 4 \cdot \int \frac{x ^{3}}{4} d x)]_{1}^{e}

\int \frac{x ^{3}}{4} d x = \frac{x ^{4}}{16}

= [\frac{1}{4} (x^{4} ln (x) - 4 \cdot \frac{x ^{4}}{16})]_{1}^{e}

4 \cdot \frac{x ^{4}}{16} = \frac{x ^{4}}{4}

= [\frac{1}{4} (x^{4} ln (x) - \frac{x ^{4}}{4})]_{1}^{e}

Compute the boundaries:

\frac{3 e ^{4}}{16} + \frac{1}{16}

= \frac{3 e ^{4}}{16} + \frac{1}{16}

Simplify

= \frac{3 e ^{4} + 1}{16}

\frac{\partial}{\partial z} (x y + z^{3} x - 2 yz)

\frac{d}{d x} (ln (\frac{3 x + 4}{3 x - 4}))

\int sin (t) 2 t d t

\int \frac{sin ( 2 x )}{e ^{x}} d x

l a pl a ce t r an s f or m 2 sin (2 t)

What is the integral from 1 to e of x^3*ln(x) ?
The integral from 1 to e of x^3*ln(x) is (3e^4+1)/(16)